Let $h$ be a polynomial function and let $h'$, its derivative, be defined as $h'(x)=(x+2)(x+1)(x-3)$. At how many points does the graph of $h$ have a relative minimum ? Choose 1 answer: Choose 1 answer: (Choice A) A None (Choice B) B One (Choice C) C Two (Choice D) D Three
Explanation: We can find the relative extrema (i.e. minima and maxima) of $h$ by looking for the intervals where its derivative $h'$ is positive/negative. A function can only change its direction from increasing to decreasing and vice versa between its critical points and the points where the function itself is undefined. We are given that $h'(x)=(x+2)(x+1)(x-3)$. $h'(x)=0$ for $x=-2,-1,3$. Since $h'$ is a polynomial, it's defined for all real numbers. Therefore, our critical points are $x=-2$, $x=-1$, and $x=3$. $h$ is defined for all real numbers so we only need to consider the critical points. Our critical points divide the number line into four intervals: $\llap{-}4$ $\llap{-}3$ $\llap{-}2$ $\llap{-}1$ $0$ $1$ $2$ $3$ $4$ $x< \llap{-}2$ $\llap{-}2<x< \llap{-}1$ $\llap{-}1<x<3$ $x>3$ Let's evaluate $h'$ at each interval to see if it's positive or negative on that interval. Interval $x$ -value $h'(x)$ Verdict $x<-2$ $x=-3$ $h'(-3)=-12<0$ $h$ is decreasing $\searrow$ $-2<x<-1$ $x=-\dfrac32$ $h'\left(-\dfrac32\right)=\dfrac{9}{8}>0$ $h$ is increasing $\nearrow$ $-1<x<3$ $x=0$ $h'\left(0\right)=-6<0$ $h$ is decreasing $\searrow$ $x>3$ $x=4$ $h'(4)=30>0$ $h$ is increasing $\nearrow$ Now let's look at the critical points: $x$ Before After Verdict $-2$ $\searrow$ $\nearrow$ Minimum $-1$ $\nearrow$ $\searrow$ Maximum $3$ $\searrow$ $\nearrow$ Minimum Now we can see that $h$ has two relative minima.